Implement hyperexponentiation/tetration without the use of '^' [closed]

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The Challenge

Implement tetration (aka Power Tower or Hyperexponentiation) with the least amount of characters.

The Conditions

• Don't use the 'power' operator or its equivalents (such as pow(x,y), x^y, x**y, etc.)
• Input given as: x y (separated by a space)
• x is exponentiated by itself y times.
• Your method must be able to compute at least 4 3 (4 exponentiated by itself 3 times)

The Scoring

• Lowest score wins: (# of characters)
• Bonus deduction if you do not use the multiplication operator (-5 points).
• No Speed/Memory requirements. Take as long as you want.

Examples

x, 0 -> 1
2, 2 -> 2^2 = 4
2, 4 -> 2^(2^(2^2)) = 65536
4, 3 -> 4^(4^4) = 4^256 =わ 13407807929942597099574024998205846127479365820592393377723561443721764030073546976801874298166903427690031858186486050853753882811946569946433649006084096

Open to suggestions/alterations/questions

• Peter Taylor: One alteration which I think is fairly important is to replace "* operator" with "multiplication operator". In GolfScript * is multiplication in some contexts, but it's also the simple looping operator: {block}N* is equivalent to C-style for(i=0;i<N;i++){block}. The tricky edge case is string/array multiplication ('a'3* gives 'aaa'), but that's unlikely to be an issue given that an array of 4***3 elements will overflow RAM.
• Peter Taylor: Also worth adding a test for the edge case x 0 => 1. My original solution didn't handle that case.
• user unknown: The penalty for using multiplication is way too low. (:=bonus for not using it). I made a solution which didn't used it, and had to replace it to avoid stack overflows, and gained a 7 char win for a 5 char bonus loss.
• MrZander: @EngineerToast I posted this golf 4 years before the one you linked...
• Simply Beautiful Art: The conditions and scoring are kind of strange. You don't allow the use of power operations? Or you do allow them, but they are a +10 point bonus?

Answer 1

J, score is 7 (12 chars - 5 points for avoiding multiplication)

+/@$/@$~/@$~

usage:

 4 +/@$/@$~/@$~ 3
1.34078e154
t=.+/@$/@$~/@$~ NB. define a function
 4 t 3
1.34078e154
 2 t 2
4

Just few nested folds:

• Using multiplication it would be */@$~/@$~
• Using power it would be ^/@$~ where $~ creates array, / is a fold function.

• Gareth: Nicely done. (pad)
• Art: @Gareth Thanks, but was does pad mean here? Sorry, English is not my mother tongue.
• Gareth: My message was too short so I needed to pad it out. :-)
• Jonah: Could you get pentation just by providing one more @$~ in the conjunction?
• hyperneutrino: @Jonah You'd need the /, but yes. you're just folding as many times as needed over the nested folding function.

Answer 2

Haskell, (削除) 87 (削除ここまで) 85 - 5 == 80 (削除) 82 (削除ここまで)

import Data.List
t x=genericLength.(iterate(sequence.map(const$replicate x[]))[[]]!!)

Uses neither of exponentiation, multiplication or addition (!), just list operations. Demonstration:

Prelude> :m +Data.List
Prelude Data.List> let t x=genericLength.(iterate(sequence.map(const$replicate x[]))[[]]!!)
Prelude Data.List> t 2 2
4
Prelude Data.List> t 2 4
65536
Prelude Data.List> t 4 3

...
ahm... you didn't say anything about performance or memory, did you? But given enough billions of years and some petabytes of RAM, this would still yield the correct result (genericLength can use a bigInt to count the length of the list).

• MrZander: I trust you will have an answer for me by 3012? ;)
• ceased to turn counterclockwis: I'll need some help from Moore's law, but that given I might.

Answer 3

GolfScript, (削除) 15 (削除ここまで) 18 chars

~])*1\+{[]+*{*}*}*

Yes, one of the *s is a multiplication operator (exercise: which one?) so I don't qualify for the 5 char bonus. Still, it's just barely shorter than Peter's solution.

This earlier 15-char version is otherwise the same, but produces no output when the second argument is 0. Thanks to r.e.s. for spotting the bug.

~])*{[]+*{*}*}*

• r.e.s.: This produces fatal errors, e.g. with "2 3" ~])*{[]+*{*}*}*.
• Peter Taylor: @r.e.s., it produces the correct answer for me.
• Ilmari Karonen: @r.e.s.: It assumes that there's nothing else on the stack but the input. If you want to provide the input in-line like in your example, first use ; to remove the actual input string that the interpreter puts on the stack at start-up. Or just prepend a [ to the code: both ;"2 3" ~])*{[]+*{*}*}* and "2 3" [~])*{[]+*{*}*}* work fine for me.
• r.e.s.: (+1) Thanks! Those variations work and solve a mystery for me. The tutorial says "You don't have to pipe input in, but if you don't, it will not prompt for input, instead it will assume there is no input." So I've been using just ruby golfscript.rb my_script.gs on the command line, without knowing that it causes something ("", apparently) to be on the stack before the script is run -- which sometimes works, sometimes not. (Also, with echo 2 3 | ruby golfscript.rb my_script.gs, your program does work as-given.)
• r.e.s.: ideone.com/547LG

Answer 4

J, (削除) 16 (削除ここまで) (削除) 19 (削除ここまで) 12 characters

*/@$~/1,~$~/

or as a verb (17 characters):

h=:[:*/@$~/1,~$~/

usage:

 h 2 4
65536

or taking input from keyboard ((削除) 24 (削除ここまで) (削除) 27 (削除ここまで) 20 characters):

*/@$~/1,~$~/".1!:1]1

with thanks to FUZxxl for pointing out my stupidity. :-)

Explanation:

J is read from right to left, so using 2 4:

/ is used to insert the verb $~ between each pair of items in the list. $~ takes the left item and shapes it $ using the right item (the ~ reverses the arguments) - so this would be equivalent to 4 $ 2 which gives you a list of 2s which is four items long 2 2 2 2.

Now we append 1 to the list 1,~ and then do the same thing again; / insert a verb */@$~ between each pair of items in the list. This verb starts in the same way $~ but this time it / inserts a * between each item of the newly generated list. The @ just makes sure that the */@$~ works as one verb instead of two. This gives 2 multiplied by itself enough times to be equivalent to 2^4.

J's vocabulary page - I find solving problems with J fun just because of the different way it sometimes does things.

Adding one further iteration to remove the * operator has 2 problems

• (削除) It comes out at 17 characters (+/@$~/,@$~/1,~$~/) which, even with the -5 bonus, is too long (削除ここまで)
• It runs out of memory if the number gets too large so doesn't meet the requirement of being able to calculate 4 3

• MrZander: Could you provide an explanation? This looks interesting.
• Gareth: @MrZander I've edited my answer to add an explanation.
• MrZander: Not sure if I have a better understanding or more confusion, but thanks haha.
• Peter Taylor: The explanation implies that the whole thing is doing exponentiation rather than tetration. Which of us is missing something?
• Gareth: @PeterTaylor I suspect my explanation is not very clear. If it was doing tetration I would have just used ^/]$[ which creates the list 2 2 2 2 and sticks the exponentiation operator between them. What this is doing is going a step further and doing the exponentiation by repeated multiplication.

Answer 5

GolfScript (24 chars - 5 = 19 points)

~1円{1{0{+}?}?}{@\+@*}:?~

is insanely slow.

(or 20 chars)

~1円{1{*}?}{@\+@*}:?~

is much faster.

• mellamokb: Since GolfScript is a Ruby program, we can test on ideone :) ideone.com/GTIfP. I've also emailed ideone suggesting they add support for GolfScript.
• Peter Taylor: @mellamokb, it'll be nice if they do add it, but I'm not too optimistic because their stated policy is to add languages which are supported by their distro.
• mellamokb: I read that too... but since they support Ruby, and GolfScript is just a Ruby program, it should be easy :) Just create a bash script that passes in the parameters.

Answer 6

Python, 70

This uses nested eval calls, eventually producing a string "a*a*a*a...*a" which gets evaluated. Almost half of the score is wasted on getting the arguments... though I've noticed that a few other solutions don't bother with that.

a,b=map(int,raw_input().split())
exec"eval('*'.join('a'*"*b+'1'+'))'*b

• st0le: If we assume the arguments are comma sepearated, you could use input() or use eval(raw_input()) Cheers
• boothby: @st0le, please read the question
• flornquake: Nice one. The second line can be golfed even more: exec"eval('a*'*"*b+'1'+"+'1')"*b

Answer 7

Br**nfuck, 128-5=123 bytes

+<<+<<,<,[>[>+>>+<<<-]>[<+>-]>[>[>>+>+<<<-]>>>[<<<+>>>-]<<[>[>+>+<<-]>>[<<+>>-]<<<-]>[-]>[<<+>>-]<<<<-]>>[<<+>>-]+<[-]<<<<-]>>>.

Input is in the form of characters with code points of the numbers desired as inputs. Output is the same.

An explanation is (削除) coming when I have the time (削除ここまで) below. Do I get bonus points for not using exponentiation, multiplication, OR even addition?

Cell 3 (0-indexed) is the running total x.
This calculates the nth tetration of a.
+<<+<<,<, Initialize tape with [n, a, 0, 1, 0, 1]
[ While n:
 >[>+>>+<<<-]>[<+>-] Copy a 3 cells to right: [n, a, 0, x, a, 1]
 >[ While x:
 >[>>+>+<<<-]>>>[<<<+>>>-] Copy a 2 cells to right: [n, a, 0, x, a, 1, a, 0]
 <<[>[>+>+<<-]>>[<<+>>-]<<<-] Cell 7 = prod(cell 5, cell 6)
 >[-]>[<<+>>-]<<<<-] Move this value to cell 5. End while.
 >>[<<+>>-]+<[-]<<<<-] Update x to result of exponentiation. End while.
>>>. Print the result!

This works (tested) for x 0, 0 x, x 1, 1 x, x 2, 2 3, and 2 4. I tried 3 3, but it ran for several hours without finishing (in my Java implementation—probably not optimal) (EDIT: in @Timwi's EsotericIDE [It's great! Y'all should try it] as well. No luck.). In theory, this works up to the cell size of the specific implementation.

• kepe: "Br**nfuck" Yes "brain" is a very offensive word xD. sorry I needed to

Answer 8

Scala:110

type B=BigInt
def r(a:B,b:B,f:(B,B)=>B):B=if(b>1)f(a,r(a,b-1,f))else a
def h(a:B,b:B)=r(a,b,r(_,_,r(_,_,(_+_))))

ungolfed:

type B=BigInt
def recursive (a:B, b:B, f:(B,B)=>B): B = 
 if (b>1) f (a, recursive (a, b-1, f)) 
 else a
recursive (2, 3, recursive (_, _, recursive (_, _, (_ + _))))

explanation:

type B=BigInt
def p (a:B, b:B):B = a+b
def m (a:B, b:B):B = if (b>1) p (a, m (a, b-1)) else a
def h (a:B, b:B):B = if (b>1) m (a, h (a, b-1)) else a
def t (a:B, b:B):B = if (b>1) h (a, t (a, b-1)) else a

plus, mul, high(:=pow), tetration all work in the same manner. The common pattern can be extracted as recursive method, which takes two BigInts and a basic function:

def r (a:B, b:B, f:(B,B)=>B):B = 
 if (b>1) f(a, r(a, b-1, f)) else a
r (4, 3, r (_,_, r(_,_, (_+_))))

The underlines are placeholder for something which gets called in this sequence, for example the addition plus(a,b)=(a+b); therefore (+) is a function which takes two arguments and adds them (a+b).

unfortunately, I get issues with the stack size. It works for small values for 4 (for example: 2) or if I reduce the depth for one step:

def h(a:B,b:B)=r(a,b,r(_,_,(_*_))) // size -7, penalty + 5
def h(a:B,b:B)=r(a,b,r(_,_,r(_,_,(_+_)))) 

The original code is 112 characters and would score, if valid, 107. Maybe I find out how to increase the stack.

The expanded algorithm can be transformed to tailrecursive calls:

type B=BigInt
def p(a:B,b:B):B=a+b
import annotation._
@tailrec
def m(a:B,b:B,c:B=0):B=if(b>0)m(a,b-1,p(a,c))else c
@tailrec
def h(a:B,b:B,c:B=1):B=if(b>0)h(a,b-1,m(a,c))else c
@tailrec
def t(a:B,b:B,c:B=1):B=if(b>0)t(a,b-1,h(a,c))else c

The tailrecursive call is longer than the original method, but didn't raise a stackoverflow in the long version - however it doesn't yield a result in reasonable time. t(2,4) is fine, but t(3,3) already was stopped by me after 5 min. However, it is very elegant, isn't it?

// 124 = 119-5 bonus
type B=BigInt
def r(a:B,b:B,c:B,f:(B,B)=>B):B=if(b>0)r(a,b-1,f(a,c),f)else c
def t(a:B,b:B)=r(a,b,1,r(_,_,1,r(_,_,0,(_+_))))

And now the same as above: use stinky multiplication (we even profit while rejecting the bonus of 5, because we save 7 characters: win=4 chars:)

// 115 without bonus
type B=BigInt
def r(a:B,b:B,c:B,f:(B,B)=>B):B=if(b>0)r(a,b-1,f(a,c),f)else c
def t(a:B,b:B)=r(a,b,1,r(_,_,1,(_*_)))

invocation:

timed ("t(4,3)")(t(4,3)) 
t(4,3): 1
scala> t(4,3)
res89: B = 13407807929942597099574024998205846127479365820592393377723561443721764030073546976801874298166903427690031858186486050853753882811946569946433649006084096

runtime: 1ms.

Answer 9

Python, 161 - 5 (no * operator) = 156

r=xrange
def m(x,y):
 i=0
 for n in r(y):i+=x
 return i
def e(x,y):
 i=1
 for n in r(1,y+1):i=m(i,x)
 return i
def t(x,y):
 i=1
 for n in r(y):i=e(x,i)
 return i

invoke:

t(2, 4)

• Peter Taylor: Is multiplication by iterated addition really fast enough to evaluate 4***3?!
• Blazer: @PeterTaylor yes? it completes in less than a second for me
• Peter Taylor: Wow. The equivalent GolfScript version takes aaaaaaages.
• Peter Taylor: As in, I've left it running overnight and it still hasn't finished.
• Jack Brounstein: Six years later, you can also save some bytes by replacing your m function with m=lambda x,y:sum(x for _ in r(y))

Answer 10

Perl, 61 chars

here's a bizarre one

sub t
{
 ($x,$y,$z)=@_;
 $y>1&&t($x,$y-1,eval$x."*$x"x($z-1||1))||$z
}

usage:

print t(2,4,1)

• ardnew: an incorrect one too

Answer 11

Mathematica, (削除) 40 (削除ここまで) 33

This doesn't quite conform to the rules but it is not in contention for the shortest code anyway, and I hope that it will be of interest to someone.

m@f_:=Fold[f,1,#2~Table~{#}]&;
m[m@Sum]

This builds a "tetration" function when it is run, but the arguments must be given in reverse order. Example:

m[m@Sum][3, 4]

1340780792994259709957402499820584612747936582059239337772356144372176 4030073546976801874298166903427690031858186486050853753882811946569946 433649006084096

• DavidC: Would you explain the code? Or display results on symbols rather than numbers? I notice that Fold[g, 1, #2~Table~{#}] &[3, 4] will produce g[g[g[1, 4], 4], 4] for instance.
• Mr.Wizard: @David m[Times] produces Fold[Times, 1, Table[#2, {#1}]] &, which is a power function: m[Times][5, x] ---> x^5; the same method is used for this new power function to produce a tetration function. Logically one could start with Plus but that fails almost immediately.
• DavidC: To eliminate Times, try this: t[h_, n_] := Sum[h, {i, n}]. Then run m[m@t][3, 4].
• Mr.Wizard: @David, yes, that should work, but not for Code-Golf. ;-) (BTW you could write Sum[h, n].)
• DavidC: Look at the scoring rules. You save 9 points by not using Times. The total score is still not better than yours but getting closer.

Answer 12

Haskell: (削除) 58 (削除ここまで) 51 chars, with or without multiplication.

i f x 1=x;i f x n=f$i f x$n-1
t=i(\o n->i(o n)n)(+)4

Ungolfed:

bump op n a = iterate (op n) n !! (fromIntegral $ a-1)
tetrate = iterate bump (+) !! 3

Shorter definition comes from inlining "bump", and defining a custom version of "iterate". Unfortunately the result is impossibly inefficient, but starting with (*) instead of (+) gives decent speed. In ghci:

Prelude> let i f x 1=x;i f x n=f$i f x$n-1
(0.00 secs, 1564024 bytes)
Prelude> let t=i(\o n->i(o n)n)(*)3
(0.00 secs, 1076200 bytes)
Prelude> t 4 3
13407807929942597099574024998205846127479365820592393377723561443721764030073546
976801874298166903427690031858186486050853753882811946569946433649006084096
(0.01 secs, 1081720 bytes)

Answer 13

Ruby (削除) 66 (削除ここまで) 59 characters

def e(x,y)
r=1
(1..y).each{t=x
(2..r).each{t*=x}
r=t}
r
end

• r.e.s.: Unfortunately, this script does not produce the correct output (1) when the second input number is 0; rather, e(x,0) returns the value of x.
• Cristian Lupascu: @r.e.s. you are right. I fixed the code. Thanks!

Answer 14

Rockstar, 233 bytes

Rockstar is a quite new computer programming language, designed for creating programs that are also song lyrics. Rockstar is heavily influenced by the lyrical conventions of 1980s hard rock and power ballads.

The solution takes two inputs, one per each line (string manipulation is very limited). It uses multiplication operator, so no bonus to apply for :/

P takes A,B
N is 1
C is 1
While C is as low as B
Let N be N of A
Build C up
Give back N
T takes E,F
If F is empty
Give back 1
Else
Knock F down
Let G be T taking E,F
Give back P taking E,G
Listen to X
Listen to Y
Say T taking X,Y

Ungolfed, here's a more lyric version (comments are in parentheses, which are standard comments in Rockstar). Variable names may take up more words, and all words count ('my power' is different from 'the power' and from 'Power')

Power takes the force and the cross (function def, two parameters: 'the force', 'the cross')
 Build my arm up ('my arm' := 1) (literally: 'my arm'++, but my arm was initialized with 0)
 Build my faith up ('my faith' := 1)
 While my faith is as weak as the cross (While loop, condition: 'my faith' <= 'the cross')
 Let my arm be the force of my arm ('my arm' := 'the force' * 'my arm' )
 Build my faith up ('my faith' ++ )
 (empty line to terminate While block)
 Give back my arm (return 'my arm')
 (empty line to terminate function block)
Tower takes the force and the enemy (function def, two parameters: 'the force', 'the enemy')
 If the enemy is nowhere (If, condition: 'the enemy' == 0) (literally: 'the enemy' == NULL, but NULL is considered 0 in arithmetics)
 Build Victory up (Victory := 1, this is a bit diffeerent from the golfed version)
 Give back Victory (return Victory)
 Else (else)
 Knock the enemy down ('the enemy' -- )
 Let the defense be Tower taking the force and the enemy ('the defense' := Tower ('the force', 'the enemy'))
 Give back Power taking the force and the defense (return Power ('the force', 'the defense'))
 (empty line to terminate If block)
 (empty line to terminate Function block)
Listen to us (input: 'us')
Listen to your enemy (input: 'your enemy')
Scream Tower taking us and your enemy (output: Tower('us', 'your enemy'))

Answer 15

JavaScript (Node.js), 41 bytes

Have some BigInts

a=>g=b=>b--?(P=n=>n--?a*P(n):1n)(g(b)):1n

Try it online!

Answer 16

Jelly, 9 bytes - 5 (no multiplication) = 4

xxxS\/\@/

Try it online!

Jelly, 6 bytes - 5 (no multiplication, but cheats by using the list product atom) = 1

xxP\@/

Try it online!

Answer 17

Python, 112 chars

The numbers should be the 1st and 2nd argument: python this.py 4 3
** operator not used.
* used. It's quite trivial to implement, exactly like **, but costs more than 5 chars.

import sys
p=lambda y:y and x*p(y-1)or 1
t=lambda y:y>1 and p(t(y-1))or x
x,y=map(long,sys.argv[1:])
print t(y)

• user unknown: How do I use the code to compute 4 3? And, just of curiosity: Have you tried to implement * that way, and to compute 4 3 then?
• ugoren: @userunknown, Input is by parameters. I added an explanation to the answer. I didn't try to add the * implementation, I believe the recursion depth would be too large for 4 3.

Answer 18

C, (削除) 117 (削除ここまで) (削除) 105 (削除ここまで) 99 chars

EDIT: Merged the two functions p and r into one, saving some chars.
Of 99 chars, 52 do the actual calculation (including variable definitions). The other 47 are for handling input and output.
(削除) BUG: Badly handles powers of 0 (e.g. 0 2). Should find a minimum cost fix. (削除ここまで) This isn't a bug, I forgot that 0 2 is undefined.

Successfully handles 4 3, and even gives an exact result. However, can be inaccurate for some smaller numbers.
Prints the number with a trailing .000000.

x,y,z;
double R(){return--y?!z?y=R(),R(z=1):x*R():x;}
main(){
 scanf("%d%d",&x,&y);
 printf("%f\n",R());
}

• mellamokb: Looks like 118 chars to me: ideone.com/9D5SU
• Sir_Lagsalot: Testing this with 4 3 is only accurate to about 18 places, double doesn't have nearly enough precision to support an exact representation.
• ugoren: @Sir_Lagsalot, double has more than enough precision for 4^256. It only has one significant digit.
• Sir_Lagsalot: Ah good point, I wasn't thinking in binary. Does it actually print out the exact value for you? It gets truncated after the first 18 or so decimal digits on my machine, but I'm willing to accept that's system specific.
• mellamokb: @Sir_Lagsalot: See the ideone link I provided. It prints out the whole number.

Answer 19

Factor, 187 characters

USING: eval io kernel locals math prettyprint sequences ;
IN: g
:: c ( y x o! -- v )
o 0 = [ x y * ] [ o 1 - o!
y x <repetition> 1 [ o c ] reduce ] if ;
contents eval( -- x y ) swap 2 c .

Before golf:

USING: eval io kernel locals math prettyprint sequences ;
IN: script
! Calculate by opcode:
! 0 => x * y, multiplication
! 1 => x ^ y, exponentiation
! 2 => x ^^ y, tetration
:: calculate ( y x opcode! -- value )
 opcode 0 = [
 x y *
 ] [
 ! Decrement the opcode. Tetration is repeated exponentiation,
 ! and exponentiation is repeated multiplication.
 opcode 1 - opcode!
 ! Do right-associative reduction. The pattern is
 ! seq reverse 1 [ swap ^ ] reduce
 ! but a repetition equals its own reverse, and 'calculate'
 ! already swaps its inputs.
 y x <repetition> 1 [ opcode calculate ] reduce
 ] if ;
contents eval( -- x y ) ! Read input.
swap 2 calculate . ! Calculate tetration. Print result.

I did not remove the multiplication operator *. If I did so, then I would need to add some logic expressing that the sum of an empty sequence is 0, not 1. This extra logic would cost more than the -5 bonus.

---

Rule breaker, 124 + 10 = 134 characters

USING: eval kernel math.functions prettyprint sequences ;
contents eval( -- x y ) swap <repetition> 1 [ swap ^ ] reduce .

This program has a lower score, but the exponentiation operator ^ breaks the rules. The rules say "(# of characters) + (10 * (# of 'power' operators))", so I applied the +10 penalty. However, the rules also say "Don't use the 'power' operator", so any program taking this penalty does break the rules. Therefore, this program of 134 characters is not a correct answer, and I must present my longer program of 187 characters as my answer.

Answer 20

Haskell 110 - 5 = 105

Tetration Peano Style. This is the most insanely slow solution possible, just a warning, but also avoids even addition.

data N=Z|S N
a&+Z=a
a&+S b=S$a&+b
_&*Z=Z
a&*S b=a&+(a&*b)
_&^Z=S Z
a&^S b=a&*(a&^b)
_&>Z=S Z
a&>S b=a&^(a&>b)

This relies on you having the patience to type out Peano numbers (and won't show the answer, If you actually want to run it, add these few lines (90 chars):

f 0=Z
f a=S$f$a-1
t Z=0
t(S a)=1+t a
main=interact$show.f.(\[x,y]->x&>y).map(f.read).words

Answer 21

Ruby, (削除) 47 46 (削除ここまで) 45

t=->x,n{r=x;2.upto(n){r=([x]*r).inject :*};r}

Answer 22

Lua: 133 chars, multiplication-less

a,b=io.read():match"(%d+) (%d+)"a,b,ba=a+0,b+0,a for i=1,b-1 do o=1 for i=1,a do o=o+o for i=1,ba-b do o=o+o end end a=o end print(o)

I was originally going to use string repetition hacks to do fake multiplication, but it likes to fail on large values. I could possibly use dynamic compilation and loadstring to make it smaller, but it's getting late here... I need sleep.

Entering "4 3" into stdin outputs:

1.3407807929943e+154

Answer 23

VBA, 90 Chars

*Perhaps the no multiplication bonus is not good enough. I think the no multiplication answer is much more interesting, but this is code golf, so it's not the best. Here's an answer without *, and a better (shorter, and better scoring) answer with it:

90 chars, no power operators, uses multiplication = 90

Sub c(x,y)
f=IIf(y,x,1):For l=2 To y:b=x:For j=2 To f:b=b*x:Next:f=b:Next:MsgBox f
End Sub

116 chars, no power operators, no multiplication bonus (-5) = 111

Sub c(x,y)
f=IIf(y,x,1):For l=2 To y:b=x:For j=2 To f:For i=1 To x:a=a+b:Next:b=a:a=0:Next:f=b:Next:MsgBox f
End Sub

NOTE: VBA has issues printing the number when the result is very large (i.e. 4, 3), but it does calculate correctly, so if, for example, you wanted to USE that number, you would be good to go. Also, even BIGGER numbers overflow (i.e. 3, 4).

Answer 24

Perl 6, 32 bytes

->\a,\b{(1,{[*] a xx$_}...*)[b]}

Try it online!

(1, { [*] a xx $_ } ... *) is a lazy sequence that generates the power tower, each element being the a list which consists of the first input parameter a replicated (xx) a number of times equal to the previous element ($_), that list then being reduced with multiplication ([*]). From that sequence we simply pluck out the b-th element.

Answer 25

Lambda calculus, 10-5

(using Church encoding and De Bruijn indeces)
λλ(1λ13)λ1

Explanation

Without De Bruijn indeces: λa,b.(b λc.ca)λc.c:

λa,b. define the anonymous function f(a,b)=
 (b apply the following function b times
 λc. the anonymous function g(c)=
 ca) apply c to a because of church encoding this is equal to a^c
 λc.c the identity function, 1 in church encoding

If you define exp_a(x)=a^x this program defines a↑↑b=exp_a^b(1) where ^b denotes function itteration.

I'm not sure if this is allowed because ca is technically equivalent to a^c how ever it is not a real built-in and only a side effect of the way integers are encoded in lambda calculus.

• Erik the Outgolfer: Hm, is there an interpreter so that I can try this? If there's no implementation of a language, then you can't use it to solve challenges here. Languages are based on their implementations here.

Answer 26

Python - 151 bytes

Python 3.0:
r=range
def exp(a,b):
 if b=0:
 a=1
else:
 r1=a
 for i in r(1,b):
 s=s*a
def tet(a,b):
 r1=0
 if b=0:
 a=1
else:
 r1=a
 for i in range(1,b):
 r1=exp(a, result) 

Here's how the new new version works.

1: Exponentiation is initialized (multiplies a by a b times), as well as tetration (in terms of our exp function).

2: a is stored in result of the tetration function.

3: Then we exponentiate a to result (as tetration does), and store that in result

4: Repeat step 3 b times.

Feel free to tell me I suck. (With huge credits to the commenters and Googology Wiki)

• rydwolf: Welcome to Code Golf! This looks like the start of a good answer, but there are a few obvious things that can be golfed (like the whitespace and variable names). You may want to check out Tips for Golfing in Python. Also, if you indent your code by four spaces here, it'll format it as code.
• user100887: Thanks for the advice! I'm going to format it now.
• user100887: I fixed a lot of issues, but I don't think this is optimal yet.
• MrZander: Welcome @StackMeter, I suggest you take a look at this post: codegolf.stackexchange.com/questions/54/… A lot of good golfing tips
• user100887: Removed the spaces and dropped about 20 bytes; thanks!

Answer 27

Wolfram Language (Mathematica), 29 bytes

(27 characters)

aNest[Nest[a#&,1,#]&,1,#]&

Try it online!

Input curried as f[x][y].

---

Without *: 35 bytes (33 characters)

aa+#&~n~0~n~1~n~1
n=Nest~Curry~3

Try it online!

Answer 28

Javascript: 116 chars

function t(i){y=i.split(' ');a=y[0];b=y[1];return+b&&p(a,t(a+' '+(b-1)))||1}function p(a,b){return+b&&a*p(a,b-1)||1}

t('4 3') Outputs:

1.3407807929942597e+154

Answer 29

Python (削除) (111) (削除ここまで) (113) no *

r=lambda x,y:(x for _ in range(y));t=lambda x,y:reduce(lambda y,x:reduce(lambda x,y:sum(r(x,y)),r(x,y)),r(x,y),1)

6***3 - 36k digits))

Upd: Have to add initial value, to fit t(X,0)=1

• MrZander: Impressive, how long did the 36k take?
• Ev_genus: 9.375 seconds including print.

Answer 30

Haskell: 88-5 chars without multiplication, 59 chars with multiplication

Without multiplication:

h x y=foldr(\x y->foldl(\x y->foldl(+)0(replicate x y))1(replicate y x))1(replicate y x)

There are probably ways that I could golf that down a little.

With multiplication:

h x y=foldr(\x y->foldl(*)1(replicate y x))1(replicate y x)

And finally, the ungolfed program:

mult x y = foldl (+) 0 (replicate x y)
expo x y = foldl (mult) 1 (replicate y x)
h x y = foldr (expo) 1 (replicate y x)

This is probably the simplest way to do this problem, which is defining multiplication as repeated addition, exponentiation as repeated multiplication, and tetration as repeated exponentiation.